Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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~Solution by MagentaCobra | ~Solution by MagentaCobra | ||
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| + | ==Solution 0.1== | ||
| + | One answer choice stands out above the rest, and that is <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}} | ||
Revision as of 20:29, 14 February 2019
Contents
Problem
Right triangle 
with right angle at 
is constructed outwards on the hypotenuse 
of isosceles right triangle 
with leg length 
, as shown, so that the two triangles have equal perimeters. What is 
?
![[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.016233639805293cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); /* draw figures */ draw((-2.,3.)--(-2.,-1.), linewidth(2.)); draw((-2.,-1.)--(2.,-1.), linewidth(2.)); draw((2.,-1.)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); /* dots and labels */ dot((-2.,3.),linewidth(4.pt) + dotstyle); dot((-2.,-1.),linewidth(4.pt) + dotstyle); dot((2.,-1.),linewidth(4.pt) + dotstyle); dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]](http://latex.artofproblemsolving.com/e/9/1/e9154dcee5b662add15f00ec3aa10f8e25240c0d.png)
Solution 1
Observe that the "equal perimeter" part implies that 
. A quick Pythagorean chase gives 
.
Use the sine addition formula on angles 
and 
(which requires finding their cosines as well), and this gives the sine of 
. Now, use 
on angle to get 
.
Feel free to elaborate if necessary.
Solution 1.5 (Little bit of coordinate bash)
After using Pythagorean to find 
and 
, we can instead notice that the angle between the y-coordinate and is 
degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point 
, we can then proceed to find the height and base of this new triangle (defined by 
where 
is the intersection of the altitude and 
) by coordinate-bashing, which turns out to be 
and 
respectively.
By double angle formula and difference of squares, it's easy to see that our answer is
~Solution by MagentaCobra
Solution 0.1
One answer choice stands out above the rest, and that is .
See Also
| 2019 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
